SA10: Truss Analysis: Method of Sections
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- เผยแพร่เมื่อ 8 ก.ค. 2024
- This lecture is a part of our online course on introductory structural analysis. Sign up using the following URL: courses.structure.education/
In addition to updated, expanded, and better organized video lectures, the course contains quizzes and other learning content.
Solution for exercise problem A: lab101.space/pdf/exercises/SA1...
Solution for exercise problem B: lab101.space/pdf/exercises/SA1...
Solution for exercise problem C: lab101.space/pdf/exercises/SA1...
Solution for exercise problem D: lab101.space/pdf/exercises/SA1...
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Thank you for a clear, concise, and consistent approach to the method of sections.
You made everything so easy to understand! Thanks a lot!
Structures have never seemed so easy! Thanks!
Your videos are awesome Dr. Structure! Keep up the good work. Thank you so much. :)
you did a hell of a good job on this video. Thanks a lot!
I wish you guys did a bunch of different subjects related to physics because it would help me immensely. I've watched two videos about this and understand it more than I ever have from my lectures. Thank you!!
+Jascha Annogui wrote: Ho do we find the force in member BN in Example 3?
Answer: This can be done in a few steps:
1. Calculate Fgh. This is done using the fbd @10:38. Taking moments about Point A, we get: 2(2) - 4 Fgh = 0. Or, Fgh = 1 kN.
2. Calculate Fmh. This is done using the fbd @9:28. Taking moments about Point B, we get: 2(4) - Fgh(4) - Fmh sin(45)(4) = 0. Since Fgh = 1, we get: Fmh = 1.414 kN.
3. Now, summing the forces in the y direction in the above fbd, we get:
Fbn + Fmh cos(45) = 2. Since Fmh = 1.414, we get: Fbn = 1 kN.
I wish my teacher would explain the topic like this... Really helpful for my course... :)
Thank you Dr. Structure this video was very helpful to me as it illustrated the method of sections very simply.
+Mayed Amwahi: asked: @5:00 why multiply 0.25 by 5?
Answer: Since we are taking moments about B, we need to multiply the 0.25 kN force by its moment arm (the shortest distance from the line of action of the force to Point B). This moment arm is the horizontal distance between A and C which is 5m.
+Jim Law asked: How can I solve for support reactions in Example 2?
To do so, consider the entire structure. It is subjected to three known forces and has three unknown support reactions. To find the reactions, write, and simultaneously solve, three equilibrium equations:
Sum of the moments about Point J must be zero.
Sum of the forces in the x direction must be zero.
Sum of the forces in the y direction must be zero.
Or,
5(10) + 3(8) + 2(2) - Ky (2) = 0
5 + 3 + Jx = 0
Jy + Ky + 2 = 0
where Ky is the vertical reaction at K and
Jx and Jy are the horizontal and vertical reactions at J.
This gives: Jx = - 8 kN, Ky = 39 kN, Jy = -41 kN.
+Vishean Veloo
Question: Hi, may I know why do you get '15' in the equation of Moment in 2.43? I dont understand that part.
Answer: The moment equation is obtained by taking sum of the moments about point H. Three forces produce non-zero moment about H: the left reaction (Ay), the applied 3kN horizontal force and the applied 4kN vertical force.
The moment arm for Ay is 20m, the moment arm for the 3kN force is 5m (the height of the truss), and the moment arm for the 4kN force is 5m. So, we get: 20(Ay) + 5(3kN) - 5(4kN) = 0.
Hence, 20 Ay + 15 - 20 = 0.
+Dr. Structure Hi I had this problem too, however it still doesn't explain how the value is 15. Is it not supposed to be 15 (distance from H) x 3 (force) which would be 60? thanks
+Kerry Aziz The moment arm for the horizontal 3kN force is not 15, it is 5. The moment arm of a force about a point is the shortest distance from the line of action of the force to that point. This would be the length of a perpendicular line from that point (here, Point H) to the line of action of the force.
What you are doing is taking the horizontal distance from Point B to Point H. That distance, 15 m, is not the moment arm for the horizontal force at B. Since it is not perpendicular to the line of action of the force.
To determine the moment arm for the 3 kN force, draw a line over the force that is an infinitely long in both directions. This is the line of action of the force. The shortest distance from point H to this line is the moment arm for the force about H. This distance is 5m, the height of the truss.
+Dr. Structure ooooh I understand now, thanks a lot!! :)
Great video, very concise! Doesn't waste time solving equilibrium equations we already know how to do and provides a broad range of examples.
Great Work. Lectures Make It sound like rocket science but you make it sound like ABC.
Very well presented in a way that even a toddler would understand lmao. Thank you for this! *subscribed*
I feel like Alfred is teaching me Trusses in Batman School
Thank you so much, Doc. Great vid! :)
DR Structure thanks for crystal clear explanation.
Nicely presented, very understandable.
Awesome.... This video will surely make me pass today!
Awesome video, very clear explanation. Thanks.
So helpful. Thank you!
Thank you so much. A lifesaver.
best structure video on youtube keeo it up
Great demonstration !
Thank you Dr. Structure it is so great videos...
Hi, may I know why do you get '15' in the equation of Moment in 2.43? I dont understand that part.
It helped me a lot! Appreciate it
extremely helpful!
You really are doctor of structure .
best explanation ever !!! u r awesome....
Thanks a lot men u are perfect. You really helped me thanks
thank you so much!
ths is jus awesome thank u
thanks you , Ill have my final exam tomorrow, and this video saved my life LOL
Great Job.. ! Awesome .... :)
ooo it's how we can easily find the force in arms of given structure...... thanks 4 the vdo sir.and also thanks for the practice questions //
All your videos are so awesome.
the explanation was awesome 👍
how can i solve the example 2 support reactions?
Hello Dr Structure, thank you very much for the great video! I have a question, though. Why is F(ac) not included in the equation of the sum of moment about D at 7:00?
Zoe L. Because the internal force in member AC does not appear in the free-body diagram (FBD). Since the force is not shown on the FBD, it does not
play a role in the equilibrium equations.
The xxternal loads applied to a structure are always shown on the free-body diagram. The force internal to a member, however, appear in the diagram only if the member is "cut" exposing the (internal) force. In the example, we cut through three truss members, but AC is not one of them. Therefore, the force in member AC has no impact on the equilibrium equation.
+Dr. Structure
In exercise problem (A) I am getting Fce as 0 kN but Fbd as 11.25 kN (T) .
please help........ Thanks.
+VIVEK SINGH Yes, your answers are correct. There should be a youtube annotation showing the correct value for Fbd.
SEARCHING EXACTLY THE TYPE OF PROBLEM TO SOLVE🧡🧡🧡
Awesome refresher
Hi .. how do i find the force BN in example 3 ?
Thanks a lot ..
Ur a legend 👏🏼
many thanks
thank you alot
best trus video on youtube
Life Saver! Just one request, maybe if you could possibly create a tutorial in analysis of structure using method of members (pin connected structures like frames). Thank you and more power :)
Thanx soo much from SAUDI ARABIA
Sultan Abbasi You are welcome!
THANK YOU
thank you
Where do you get the (2)(2) from at 10:40?
That is the moment of the 2 kN force about point A. The moment arm for the force (the shortest distance from the point of application of the force to point A is 2 m).
just awesome
When taking moment about point B in example 1, why isn't " +Fbesin45" considered? Would that distance no also be considered perpendicular to B?
Since Fbe passes through B, it creates no moment about B. Simply put, if you draw the line of action of the force and it passes through a point, the moment of the force about the point is zero.
Dr. Structure, thanks alot...I have shorter process now(only moment equilibrium eqn can be used to find all member forces through method of sections
It is awesome but got a question- at duration 13.55 of this video, in second scenario why did you not include 2KN force acting at point C and why ( Fdf *2) is negative whilst in first scenario all forces are positive..
Gurwinder singh In the second scenario we are taking sum of the moments about point G. Since the 2 kN force passes through G, it has a zero moment arm, meaning its moment about the point is zero.
The equation assumes the moments that have a clockwise rotation about G to be positive. Here, Fdf causes a counterclockwise moment about G, hence the negative sign.
I wish u taught in our university
I would like to thank you for this great explanation, but i'm not quite sure why did you multiply 0.25 by 5 (@5.00)
in example-2 can we take moment about B to find out force in member Fce???? similar to example 1. i took this but did not get wright answer.
+Nadeem Kuskiwala Not directly. The moment equation about B includes both Fce and Fcd. Unless we know Fcd already, the equation has too many unknown for determining Fce.
finding out Fgi where your working shows 5x8+3x6+2x2+Fgi x2 = 0 where do you get the 2X2 when the distance to H is 6 metres from C. also how do you know to ignore all the joints except A, B, C and Fgi is it because you do not know the forces at the other joints. thanks.
+xganh zu There is an external 2kN force applied at C, the shortest distance from C to H is 2m, not 6m. So, the moment that the force creates about H is 2x2.
The reason we ignore all the intermediate joints in writing the equilibrium equation is that there are no external forces applied at those joints. The applied loads are only at A, B and C.
ohh ok thx
Genius!
Perfect:-)
When you took your equilibrium equations for the first truss example, you get H_x to be -3kN, however it looks like you assumed to the right as positive, and if you got a negative value, why not draw the arrow pointing to the left?
+Lake Graham Sure. You can draw the force in the other direction and show its magnitude to be positive. It is correct either way.
In example 3. what if we want to find the force MB? Should we still use the method of sections? Thanks!
Yes, I would use the Method of Sections to find the force in member MB. But this involves two steps:
First, determine the force in member AB. Then, cut the truss through members AB, MB, MH and GH.
Now set sum of the moments about point H for the right substructure to zero. The moment equation contains only one unknown, the force in member MB. Solve the equation for the unknown force.
Could you explain what the equations are in exercise A? ive got Fce as 0 but Fbd as 11.25.... Thanks
I got the same answer i think they mistyped it
@5:52 im trying to figure out why we did the moments minus Fce multiplied by 5. Isnt Fce in the x direction? isnt moments focused on forces in the y direction. So why was negative Fce multiplied by 5 used? And should that be positive Fce multiplied by 5 since the force is to the right? Alsl I learned that clockwise is negative so i am so confused.
+Peta-anne D Here, we are writing the equilibrium equations for the free-body diagram consisting of joints A, B and C.
I am assuming your question(s) pertain to the third equilibrium equation:(0.25)(5) - Fce (5) = 0.
This is a moment equation, we are taking sum of the moments about joint B.
You asked: Isn't Fce in the x direction? Isn't moment focused on forces in the y direction?
The answer to the first question is YES, Fce is in the x direction. The answer to the second question is NO. Regardless of its direction, any force that does not pass through joint B has a moment about B.
Here, five forces are shown on the free-body diagram (Fbd, Fbe, Fce, the 3kN force and the 0.25kN force). Three of these forces (Fbd, Fbe and the 3kN force) pass through joint B, therefore their moment about B is zero. The remaining two forces (Fce and the 0.25kN force) however have non-zero moments about B.
The moment of the 0.25kN force about B equals to the magnitude of the force times its moment arm (the shortest distance from the line of action of the force to B). This distance is 5 m (the horizontal distance between A and C).
Similarly, the moment of Fce about B is Fce times the shortest distance from the line of action of the force to B (the moment arm). This distance is 5m (the distance between C and B).
As for the sign convention for moment, you can assume either clockwise or counterclockwise to be positive, it does not make any difference which direction is considered positive as long as it is done consistently.
If we assume counterclockwise is positive, we get:
(5)(Fce) - (0.25)(5) = 0.
If we assume clockwise is positive, we get:
(0.25)(5) - Fce (5) = 0.
Either way, we get Fce = 0.25 kN.
Ohhhhhhh wow!!! Great explanation thank u so much!!!
+Dr. Structure can I learn forming the moment equation from any of your other videos?
I Could vaguely understand the (Fce)(5) part.
+Taanya Sachdeva We are summing the moments about B. The 0.25kN reaction force creates a clockwise moment (we assumed clockwise to be positive). But Fce creates a counterclockwise moment, hence the negative sign.
The moment arm for the 0.25kN is 5m. This is the horizontal distance between A and B.
The moment arm for Fce is also 5m. This is the vertical distance between C and B.
So, we get: +0.25(5) - Fce(5) = 0.
Dr. Structure thank you so much :)
Answer of exercise problem A should be Fbd=11.25 kN.
Could please explain how did you get the 3 equations especially the third one 2:30 ?
Summing the moments about point H, clockwise assumed positive:
Three forces have a non-zero moment about H: The forces and their moment arms are:
Force: Ay
Moment arm: 20 m
Force: 3 kN (horizontal force at B)
Moment arm: 5 m (the shortest distance from the line of action of the force to H. This is the height of the truss).
Force: 4 kN (the vertical force applied at F)
Moment arm: 5 m (this is the distance between G and H).
Equation: (20)(Ay) + (5)(3) - (5)(4) = 0 .
The moment arm for the 3kN force is the shortest (perpendicular) distance between the force and point H.
In order to determine the distance, extend the line of action of the force indefinitely in both directions. Here the line is going to be horizontal, since the force is horizontal, going from negative infinity to positive infinity in the x direction.
The shortest (perpendicular) distance between the force and H is the same as the perpendicular distance between H and this infinitely long line.
All we need to do is to draw a line from H that makes a 90-degree angle with the line of action of the force. Basically, that would be the dimension line, labeled 5 m, that appear at the right side of the truss drawing @3:04
how to calculate the support reactions when we have more unknowns than equations and no hinges? ex B and D
( regardless of method of section in which may not be necessary)
+Omid Hassani Such a structure is called statically indeterminate. We need to resort to more advance techniques for their analysis such as the Force Method (a classical method) or the Displacement Method.
+Dr. Structure
Thanks for the answer! It means that they are advanced methods. My question is why when we have no clue of such methods in Elementary Structural Analysis, statically undetermined structures are subjects for the final exams and specially to calculate the support reactions?!
Would you please teach us one of those methods?; so we, Dr Structure 's students, will be able to solve such cases.
+Omid Hassani Omid, we are working on couple of lectures covering the analysis of statically indeterminate beams and trusses. Unfortunately, they are not going to be ready before this semester's final exam.
our instructor taught us that in method of sections we can only bisect 3 members but on the book i read it says the same but some of the examples bisect 4 members//confused
Thanks for the question.
There are generally no rules that prevent us from cutting through more than, or less than, three members.
In this method we use the conditions of static equilibrium (the equilibrium
equations) to find unknown member forces. Since there are three equilibrium equations (assuming a planar truss) then we cannot solve for more than three unknown forces at a time. Under the assumption that you wish to determine the axial force in ALL the cut members, then we should not cut through more than three members. If we cut through say 4 members, then we end up having 4 unknowns but only 3 equations.
But what if we don't necessarily want to calculate all the unknown forces that
result from cutting the truss? What if we want to calculate the force in a single member only? Do we still need to cut through three members? The answer is no. The number of members that we cut through is rather irrelevant here. What is relevant? That the cut enables us to find the unknown force using one of the equilibrium equation (generally the moment equation).
More specifically, say we cut through N members exposing N unknown forces one of which being the force we wish to determine. If all the forces except our target force pass through a single point, then we can solve for the target force by writing the moment equilibrium equation about that point. That is since most of the forces pass through the point, their moment about the point would be zero. The only non-zero term in the moment equation would be due the target force. We can then use the equation (which contains only one variable) to solve for the target member force. This is how a K-truss generally analyzed using the Method the Sections.
Dr. Structure thanks a lot.
thank you for the video. I just want to ask, fir the questions given at the end of the video, why is the value for F(ce)=0 for question A). Your reply is greatly appreciated
+SuperTaiko95 Cut the truss vertically through members BD, BE and CE. Now, consider the left segment of the structure. Its free-body diagram consists of four forces: the vertical reaction at the roller support, Fce, Fbe and Fbd.
Notice that three of these forces (the support reaction, Fbd and Fbe) pass through point B. So, if we take sum of the moments about B, we get:
4 Fce = 0
This means for the equilibrium equation to be satisfied, Fce has to be zero.
Ok. That makes sense. But what if I was to consider the right hand-side of the structure? Do I have to take into account the reaction forces at point E as well as the other forces?
+SuperTaiko95 Yes, we can take the right substructure. But then its free-body diagram would consist of Fbd, Fbe, Fce and the horizontal and vertical reactions at E as well as the applied loads at G and H. Now we have five unknown forces meaning we need to do a bit of calculations before we can show that Fce is zero.
+Dr. Structure My answer is Fce=0 and Fbd=-11.25. But the correct answer is Fbd=1.25 Kn. Why? please healp me.
+kazuto kirigaya Yes, the correct answer is 11.25 kN, the member is in tension. There is a (youtube annotation) note placed on the video clip @15:10 that shows the correct value.
I'm confused about the 2nd example where you solve for DF. I thought that the moment arm of a force about a point is the shortest distance from the line of action of the force to that point... So then why are we multiplying 5 kN by 6m (vertical) instead of 2m (horizontal) and same for 3kN why is it 4m instead of 2m? I also don't understand why Fdf(2) is negative in this equation... Please help!
Yes, the moment arm is the shortest distance from the line of action of the force to the point. Here, the force is 5 kN and its line of action is a horizontal line that passes through point A. The point that we are taking moments about is G. To determine the moment arm (the shortest distance from G to the horizontal line that passes through A), draw a vertical line from G until it intersects the line of action of the 5 kN force. This vertical line is the shortest distance from the (horizontal) line of action of the force to G. The height of this line is 6m ( 2 + 2 + 2) since each vertical segment of the truss is 2m in height. So, the moment of the 5kN force about G is (5 kN)(6 m).
Moment arm is *always* perpendicular to the line of action of the force. So, extend the force by drawing an infinite line that passes through its tail and head. That is the line of action of the force. Now, identify the point that you are taking the moments about, and from that point draw a line that makes a 90-degree angle with the line of action of the force. The length of that line is the moment arm.
So, the line of action of the 3 kN force is a horizontal line that passes through points B and C. The moment arm about point G is the vertical line GEC since that line makes a 90-degree angle with line BC. Length of GEC is 4m. So the moment of 3 kN force about G is (3 kN)(4 m).
As for the sign, the 5 kN force creates a clockwise moment around G. The 3 kN force also causes a clockwise rotation around G. But, Fdf causes a counterclockwise rotation. If we take clockwise rotation as positive, then counterclockwise would be considered negative. This gives us:
+ (5)(6) + (3)(4) - (Fdf)(2) = 0
How do we know if the moment of a force around a point is clockwise or counterclockwise? Imagine you have a piece of string. You attach one end of it to G and the other end of it to the head of the force vector. Taking G as the center of a circle, and the string as the radius of the circle, then make the arrow to rotate around G with the same radius. This circular motion of the force is either clockwise or counterclockwise depending on the direction of the arrow. In the case of the 5 kN force, the string is going to turn clockwise. In the case of Fdf, the string is going to turn counterclockwise.
Ah! Now I get it! Thank you!!
at 5.08, why did u take fce as negative, shouldn't it be positive since its going to the right??
When writing the moment equilibrium equation, we need to assume either clockwise or counterclockwise as the positive direction. In this case, we assumed clockwise moment to be positive.
We are writing the sum of the moments about point B. The support reaction (0.25 kN) created a clockwise moment of (0.25 kN)(5 m) about point B. And Fce creates a counterclockwise moment of (5 m) (Fce). So, the moment equilibrium equation needs to be written as:
+(0.25)(5) - (Fce)(5) = 0
Where did the "20Ay +15 -20=0" came from?☺ (In the beginning of the example)
The equation represents the sum of the moments about H. There are three forces that create a moment about H:
The vertical reaction at A. Its moment about H is: 20 Ay (clockwise)
The horizontal (3 kN) force at B. Its moment about H is: (3)(5) = 15 (clockwise)
The vertical (4 kN) force at F. Its moment about H is (4)(5) = 20 (counterclockwise).
Adding these three terms two of which is clockwise and one counterclockwise gives us: 20Ay + 15 - 20 = 0
They look so easy here.
Hey Dr. Structure. Thank you for the great video it was really helpful. I have a question regarding example 3. How do you compute the reactions for BN and NH? Or is it since the trusses are symmetrical then the reactions would be zero? Thank you in advance for your reply.
The support reactions for the truss are computed before we start cutting the truss. And, they remain the same regardless of how we cut the truss. In this case, the left and right support reactions are 2 kN each, as long as we don’t change the applied loads. That is, as long as the truss is subjected to a downward 4 kN force at joint I, the reactions at joints F and L remain the same.
Hi, I'm not very good in this mateer, and I would like to know if it was possible to have a solution for the problem "A" please!
Thanks
+Darkozy dofus I think the solution that you are looking for was discussed in this comment section, take a look and see if you can find the answer to your question.
how to solve exercise b? i tried cutting it in ab but needed value for ay.
You don't need Ay. Cut AB, take the right substructure (that would be everything except A and the lower part of AB), then sum the moments about C. Two forces cause moment about C in that substructure, the applied load and the force in AB. That means the moment equation has only one unknown in it, the force in AB. Solve the equation for the unknown force.
very easy way to represent☺
I have a question about 2nd equilibrium equation @2:38 ... it's Summation F(X) and hence equation i was expecting was -H(X)+3 = 0 , but the equation says H(Y) ! Can someone explain. Thanks
Yes, the correct equation is Hx + 3 = 0, NOT Hy + 3. The correction was made using TH-cam annotation which you should be able to see on a desktop. Unfortunately, TH-cam annotations don't show up on mobile devices.
super super................
i hope there is a video of solution of the excercise problems
Hi Dr. Structure,@7:11 you wrote the equation of moment about point D. According to you, the forces that make moment at point D are: 5kN at point A, 3kN at point B, 2kN at point C and the force Fce in the line of action CE. My question though is: why the force Fcd does not make moment at point D too? Because the components of that force would be Fcd (x) and Fcd (y), in a distance of 2 and 2 meters, respectively. Then our equation of moment would be a little bit different, adding those two forces.
Amarildo Kuhn Okay, let's write the moment of Fcd about point D. Fcd is an inclined force, it has an x component and a y component. Since the angle of inclination is 45 degrees, the component forces are:
x-component of Fcd = 0.707 Fcd
y-component of Fcd = 0.707 Fcd
Now let's take the moment of Fcd about point D. Assume clockwise moment to be positive.
The moment is: 0.707Fcd (2) - 0.707Fcd (2) = 0.
That is, the x and y components have the same moment value about D, but in opposite directions, so they cancel each other out.
Alternatively, if you examine the line of action of Fcd, you can see that it passes through point D. Therefore, its moment arm about D is zero. This means the moment of the force about D is zero.
Dr. Structure Thanks for your answer. I was really confused about this, but now it's clear. They both cancel each other.
However, let's suppose that the length of square was 3 in the x-axis and 2 in the y-axis. We would ended up finding an angle of 33.7 degrees. In this case, we would have:
x-component of Fcd = 0.83 Fcd
y-component of Fcd = 0.55 Fcd
In this assumption, even though the line of action of Fcd passes through point D, the x and y components would have different moment values about D, is that correct?
Amarildo Kuhn That is right, the components would be different if the angle is different than 45 degrees, but so would the two moment arms.
For your example we get: (0.83 Fcd)(2) - (0.55 Fcd)(3) = 0.
in problem A the support reaction C is negative. it means that the support should have a downward reaction. but it's a roller, how can a roller have a downward reaction? i dont think the truss is statically determinant in this way. please answer
Roller supports are often found in heavy structures like bridges where dead load is significantly larger than any load that could result in an uplift force at the support. In structural analysis we often calculate reaction and member forces for individual loads separately, then add them up according to design equations to determine design loads. Although the roller could be subjected to an uplift force under, say, the load due to a moving vehicle, but the overall force on the support is going to be downward when all the other loads are added to the mix.
In other types of (lighter) structural system where an uplift force could develop at a roller support, various mechanisms (e.g., sliding bearings) can be used to ensure that the support remain attached to its base.
wow, that was a good explanation. i didnt think of that.
why do we ignore the force acting on BC @4:05?
When writing the equilibrium equations we only consider the forces that appear (are drawn) on the free-body diagram including any support reactions, applied loads and the internal force in each cut truss members. If a member is not cut, its internal force is not going to show up on the free-body diagram. Here, AB, AC and BC are not cut, so they are not considered when writing the equilibrium equations.
@@DrStructure @13:06 5(8) +3(6) +2(2) +fgi(2), where did you get 2(2) value? And fgi(2) is the force multiplied by the distance 2m right?
@@aveira.a4916 There is a vertical force of 2 kN at joint C. The line of action of that force is an infinitely long vertical line that passes through joint G. The shortest distance (the moment arm) for that force, therefore is the horizontal distance between joint H and that vertical line. That distance is 2 meters. Hence, the 2 kN force creates a clockwise moment of (2 kN)(2 m) about H.
And, yes, that moment arm for Fgi is 2 m, the same distance mentioned above. It is the same distance since both 2 kN force at C and Fgi act along the same vertical line.
How would we solve if we know the lengths but only have a center force?
What is a center force? Please elaborate.
For exercise problem B I got 3.35kN by taking the moments around pt C. I calculated the distance the 2kN was from pt C by first calculating the distance DC by taking 5xCos(15)x2 then multiplied that number by the cos(30) to find the vertical distance of the force from the point of rotation. This gave a moment of 16.73kNM. Since this moment can only be cancel by the tension on beam AB I divided this moment by 5 to get 3.346kN. What am I doing wrong?
+Curt D Distance CD (the shortest distance between the applied load at Point C) is: 5 + 5 sin(45) = 8.536. Then, the moment that the 2kN force generates about C is: 2(8.536) = 17.07. Now, we can get the force in AB using equation:
5 Fab = 17.07. Or, Fab = 3.41 kN
+Dr. Structure Thank you, I made an assumption that angle BCE was 15 degrees because I though ACD was an equilateral triangle. I am clearly wrong because angle ABC is 90 not the required 120 to form such a triangle. Thank you for your quick response. I will try to assume less.
@@DrStructure but Fab is -3.416kN,isn't it????
No, the member needs to be in tension, meaning the force would be pointing away from joint B, for the sum of the moments about C to be zero. The applied load creates a clockwise moment about C, so Fab must create a counterclockwise moment about C for the sum to be zero. That means the (Fab) force arrow must be point away from B and toward A, that signifies a tensile force.
I did not understand 5fab=17.07
are you taking moments of counter clockwise as positive?
forget it ive answered my question when you said assume clockwise is positive. this makes up for the reaction forces being different
Nice Video
hi Dr.structre i am unable to get the answer for exercise problem 3..can u help me with giving some hints..
Rinil Babu Cut through members CE, CD and BD. Draw the free-body diagram for the left substructure. Three unknown member forces are present in the diagram. Two of the unknown forces pass through a point, you want to write the moment equation about that point. The equation would contain only one unknown, Fbd.....
Dr. Structure thanku i got the correct answer..its always been a tough topic for me sometime..now u made it very easy for us..a lot of thanks
where can i find the solution for the exercise.
The answers are given at the end of the lecture. Review comments for additional information about the individual problems.
Where frm can i get answers to the end problems to cross check plz do reply
We will provide the solution for each problem in pdf format within 24 hours. You will be able to find links in the video description field then.
The exercise problem solutions are now available. See links in the video description field.
is there a reason this video represents clockwise moments as positive instead of negative?
When summing moments about a point, the assumed sign convention is rather arbitrary. We can use either clockwise or counterclockwise as the positive direction.
Yeah, I guess what im asking is, why don't you follow the right hand rule?
good one
How do you know when to use cos or sin @4:50
Generally, you can use either sine or cosine. It is more a matter of preference or convenience.
Good day.
I have been searching for the solutions for this SA10 on methods of section.
Please see the comment sections, hints and solutions are presented there.
@@DrStructure Thank you
Very nice lecture. Tnx a zillion.
But will like u to solve the exercise given at the end of the lecture in SA11 METHOD OF SECTION.
Tnx